Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

0 APPENDIX
Appendix H Complex Numbers
1. (5- 6i) + (3 + 2i) = (5 + 3) + ( -6 + 2)i = 8 + ( -4)i = 8 - 4i
3. (2 + 5i)(4- i) = 2(4) +2(-i) + {5i)(4) + (5i)(- i) = 8 - 2i + 20i - 5i 2 = 8 + 18i- 5(-1)
5. 12 + 7i = 12 - 7i
= 8 + 18i + 5 = 13 + 18i
1+4i 1 +4i 3 - 2i 3 - 2i+ 12i-8(-1) 11 + 10i 11 10 .
7· 3 + 2i = 3 + 2i . 3 - 2i = 3 2 + 22 = 13 = 13 + 13 ~
1 1 1-i 1-i 1- i 1 1 .
9 -- = -- . - - = = -- = - - - t
. 1 +i 1+i 1- i 1 -(-1) 2 2 2
11 . i 3 = i 2 . i = ( - 1 )i = -i
13. yC25 = J25 i = 5i
15. 12 - 5i = 12 + 15i and 112- 15il = J12 2 + ( -5)2 = )144 + 25 = Jl69 = 13
17. - 4i = 0 - 4i = 0 + 4i = 4i and l-4il = J0 2 + ( -4) 2 = VW = 4
19. 4x 2 + 9 = 0 # 4x 2 = - 9 # X
2- _Q
- 4
21. By the quadratic formula, x 2 + 2x + 5 = 0 # x =
- 2 ± J2 2 - 4(1)(5) -2 ± v'-16 -2 ± 4i .
2 ( 1 ) = 2
· · -1 ± y'1 2 - 4(1)(2) - 1 ±A 1 · v'7
= - 2 ± 2i.
23. By the quadratic formula, z 2 + z + 2 = 0 # z =
2 ( 1 ) = 2
=
2
= -1 ± 2t.
25. For z = -3 + 3i, r = J( -3) 2 + 3 2 = 3 V2 and tan 8 = ! 3
= -1 => 8 = 3 ; (since z lies in the second quadrant).
Therefore, - 3 + 3i = 3 J2 (cos 3 4 ,. + i sin 3 ;).
27. For z = 3 + 4i, r = )3 2 + 4 2 = 5 and tan 8 = ~ => 9 = tan - l ( ~) (since z lies in the first quadrant). Therefore,
3 +4i = 5[cos(tan- 1 ~) +isin(tan- 1 1)].
29. For z = .J3 -H, r = J ( v'3) 2 + 1 2 = 2 and tan 9 = -ja => 8 = % => z = 2 (cos -if + i sin -if).
Forw=l+v'3i, r=2andtanB=v'3 => 8 =~ => -w =2(cosi + isin~).
Therefore, zw = 2 · 2 [cos ( ~ + ~) + i sin ( i + ~)] = 4 (cos i + i sin i),
z / w = H cos ( i - i) + i sin (% - i)] = cos (- %) + i sin (-f), and 1 = 1 + Oi = 1 (cos 0 + i sin 0) =>
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